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I Doubts Arising from Clausius' Inequality and the Second Law

  1. May 9, 2017 #1
    I began reading Mehran Kardar's Statistical Physics of Particles and about halfway through the first chapter, there was a discussion on the second law of thermodynamics. He makes no mention of the old tenet that 'the total entropy in the universe must always increase' (I'll refer to this as the 'Second Law'); instead, the discussion revolved entirely around heat engines and Kelvin's, Clausius', and Carnot's theorems.

    Kardar derived Clausius's Inequality using a derivation exactly similar to the one discussed in this thread. Since there was no mention of the Second Law anywhere in the derivation, I turned to the internet in hopes of better understanding it's relation to Clausius' Inequality. I quickly found this cogent derivation, which helped me out a lot but ended up raising more questions in it's wake. I'll summarise this derivation in the next paragraph and explain my doubts in the one after that.


    Assume two bodies at temperatures [itex]T_1[/itex] and [itex]T_2[/itex] ([itex]T_1 > T_2[/itex]) are exchanging an amount of heat [itex]\delta{Q}[/itex]. The second law dictates that the total entropy must always increase. This leads to the intuitive result that heat must flow from the higher temperature body to the lower temperature one, as can be verified from the following:
    [tex]\delta{S} = \frac{-\delta{Q}}{T_1} + \frac{\delta{Q}}{T_2} \geq 0[/tex]
    Now, if we apply this idea to a cyclic process on a system exchanging heat with it's environment, we get:
    [tex]\oint{dS_{system}} + \oint{dS_{environment}} \geq 0[/tex]
    The first term, [itex]\oint{dS_{system}} = 0[/itex] because [itex]S[/itex] is a state function. The second term can be written in terms of the heat exchange as seen from the point of view of the system (negative of the changes as seen from the point of view of the environment). From this we get Clausius' inequality:
    [tex]\oint{\frac{dQ}{T}} \leq 0[/tex] This completes the derivation.


    I have two problems with this derivation:
    1. My first problem is with the assumption that [itex]\oint{dS_{system}} = 0[/itex]. Let's go back to the basic definition of entropy for a second: [itex]dQ = T dS[/itex]. Using this, we get that [itex]\oint{S_{system}}[/itex] is actually [itex]\oint{\frac{dQ}{T_{system}}}[/itex]. This is zero only when the path taken is reversible, i.e., when [itex]dQ = {dQ}_{reversible}[/itex]. Only if the process is reversible can we calculate [itex]\oint{dS_{system}}[/itex] using the state function [itex]S[/itex] and conclude that it is zero. It seems that somewhere the 'actual' entropy change in the system is assumed to be the same as if the system changed states reversibly, and this shouldn't be right...
    2. Secondly, the final term left behind in Clausius' Inequality is actually the [itex]\oint{dS_{environment}}[/itex] term; so, the temperature in the denominator of the inequality is actually the temperature of the environment, and the inequality should read: [tex]\oint{\frac{dQ}{T_{environment}}} \leq 0[/tex] Now, there's no reason that we should be able to equate [itex]T_{environment}[/itex] with the system's temperature, [itex]T[/itex]; yet, in the derivation of identities like [itex]dQ \leq T dS[/itex] (follow the link; the derivation is very succinct), it seems that this is exactly what we are doing.
    Last edited: May 9, 2017
  2. jcsd
  3. May 9, 2017 #2


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    Staff: Mentor

    You need to read Grandpa Chet's Entropy Recipe.

    T is a positive quantity, so it doesn't matter which value you use for the inequality.
  4. May 9, 2017 #3

    Stephen Tashi

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    Doesn't that definition assume the equation applies to a situation where the state of the system is passing through equilibrium states ? - thus making the process reversible.

    How can we define the thermodynamic entropy of a gas not in equilibrium?

    How can we even define "the temperature" of a gas not in equilibrium?

    For example, suppose you have gas in an equilibrium state in a container and open the container so the gas begins to flow out of it. It's unclear what the phrase "the volume of the gas" would signify when the gas is flowing out. If "the volume" means the space occupied by the individual molecules then, for an ideal gas, that volume would be zero (before and after the container is opened) because molecules are "point particles".

    The problem of defining "the temperature" of the flowing gas could be approached by the thought that "temperature is the mean kinetic energy". However, a statistical mean of something is an average computed with respect to some other quantity - so we could have a "mean kinetic energy per molecule" or a "mean kinetic energy per cubic cm" etc.

    If we want temperature to be the "mean kinetic energy per cubic cm" then we have the problem of setting some sort of boundary for "the volume" of the flowing gas. If we don't set such a boundary then we average the kinetic energy over all of space and get zero.
  5. May 9, 2017 #4
    To get the entropy change of a system that has experienced an irreversible process, the first thing you need to do is establish the final state of the system, using the 1st Law of Thermodynamics. After that, you can forget entirely about the irreversible process; it is of no further use. Next, you need to devise a reversible path between the same initial and final thermodynamic equilibrium end states, and calculate the integral of dQ/T for that path. The heat Q for that path does not have to equal the heat for the irreversible path, and the reversible path does not need to bear any resemblance to the actual irreversible path. The integral of dQ/T for the reversible path will be the change in entropy. Note that, as expected, this depends only on the two end states. For a cyclic path that starts and ends at the same state, the integral will, of course, be zero. All reversible paths will give exactly the same value for the integral.

    Back to the Clausius Inequality. The temperature in the Clausius inequality for an irreversible path is not the average temperature of the system. It is the temperature at the boundary of the system, at the location where the heat flow is crossing the boundary (This is a little know fact, typically not emphasized, or even recognized, in thermodynamics textbooks and online sources). So, the Clausius inequality should really read:
    $$\Delta S_{sys}\geq \int{\frac{dQ}{T_B}}$$where ##T_B## is the temperature at the heat transfer boundary.

    As DrClaude has suggested, it would be very helpful if you read my Physics Forums Insights article, Grandpa Chet's Entropy Recipe. This should clear up all your doubts.
  6. May 9, 2017 #5

    Bam! :wideeyed:
    Okay, that was the crucial insight...everything makes sense now. :smile:

    Yes, this makes a lot of sense. Something on these lines was suggested in this article, which I found in the suggested articles on your's. BTW, the examples given in your article were really nice!

    Finally, seeing that entropy changes can only be defined for systems in equilibrium (undergoing quasi-static processes), the temperature of the reservoir and the interface can safely be assumed to be equal to the temperature of the system, right? Then the usual interpretation of the temperature [itex]T[/itex] as the system temperature follows.
  7. May 9, 2017 #6
    For the reversible path that you devise, the temperature of the reservoir(s) differs only slightly from that of the system over the entire path. If the system temperature is changing over the reversible path, you may need to use a sequence of reservoirs, differing from one another in temperature only slightly along the path. Note also that, for an ideal reservoir, irrespective of whether the path is reversible or irreversible, the boundary temperature is taken to be the reservoir temperature.
  8. May 10, 2017 #7
    Yes, I saw the series of reservoirs trick in one of the applications of your recipe.

    Thank you for your help.
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